Show that the function $f(X)=X^{-1}$ is matrix convex on $\mathbf{S}_{++}^n$.

Solution. We must show that for arbitrary $v \in \mathbf{R}^n$, the function
$$g(X)=v^T X^{-1} v$$
is convex in $X$ on $\mathbf{S}_{++}^n$. This follows from example 3.4.
4.1 Consider the optimization problem
$$\begin{array}{ll} \text { minimize } & f_0\left(x_1, x_2\right) \ \text { subject to } & 2 x_1+x_2 \geq 1 \ & x_1+3 x_2 \geq 1 \ & x_1 \geq 0, \quad x_2 \geq 0 . \end{array}$$
Make a sketch of the feasible set. For each of the following objective functions, give the optimal set and the optimal value.

Log-concavity of Gaussian cumulative distribution function. The cumulative distribution function of a Gaussian random variable,
$$f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^x e^{-t^2 / 2} d t$$
is log-concave. This follows from the general result that the convolution of two logconcave functions is log-concave. In this problem we guide you through a simple self-contained proof that $f$ is log-concave. Recall that $f$ is log-concave if and only if $f^{\prime \prime}(x) f(x) \leq f^{\prime}(x)^2$ for all $x$.
(a) Verify that $f^{\prime \prime}(x) f(x) \leq f^{\prime}(x)^2$ for $x \geq 0$. That leaves us the hard part, which is to show the inequality for $x<0$.
(b) Verify that for any $t$ and $x$ we have $t^2 / 2 \geq-x^2 / 2+x t$.
(c) Using part (b) show that $e^{-t^2 / 2} \leq e^{x^2 / 2-x t}$. Conclude that
$$\int_{-\infty}^x e^{-t^2 / 2} d t \leq e^{x^2 / 2} \int_{-\infty}^x e^{-x t} d t .$$
(d) Use part (c) to verify that $f^{\prime \prime}(x) f(x) \leq f^{\prime}(x)^2$ for $x \leq 0$.

Solution. The derivatives of $f$ are
$$f^{\prime}(x)=e^{-x^2 / 2} / \sqrt{2 \pi}, \quad f^{\prime \prime}(x)=-x e^{-x^2 / 2} / \sqrt{2 \pi} .$$

(a) $f^{\prime \prime}(x) \leq 0$ for $x \geq 0$.
(b) Since $t^2 / 2$ is convex we have
$$t^2 / 2 \geq x^2 / 2+x(t-x)=x t-x^2 / 2$$
This is the general inequality
$$g(t) \geq g(x)+g^{\prime}(x)(t-x),$$
which holds for any differentiable convex function, applied to $g(t)=t^2$ Another (easier?) way to establish $t^2 / 2 \leq-x^2 / 2+x t$ is to note that
$$t^2 / 2+x^2 / 2-x t=(1 / 2)(x-t)^2 \geq 0 .$$
Now just move $x^2 / 2-x t$ to the other side.
(c) Take exponentials and integrate.
(d) This basic inequality reduces to
$$-x e^{-x^2 / 2} \int_{-\infty}^x e^{-t^2 / 2} d t \leq e^{-x^2}$$
i.e.,
$$\int_{-\infty}^x e^{-t^2 / 2} d t \leq \frac{e^{-x^2 / 2}}{-x}$$
This follows from part (c) because
$$\int_{-\infty}^x e^{-x t} d t=\frac{e^{-x^2}}{-x}$$

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